ME 201
Engineering Mechanics: Statics
Chapter 8
Friction
Friction
 A retarding force that resists the relative movement
of two bodies in contact with each other



Always acts in oppose direction of motion – actual motion
or impending motion
Always acts parallel or tangent to surface(s) in contact
Magnitude is mainly dependent on surface roughness
(other contributing factors: temperature, molecular
adhesion, electrostatic attraction, lubrication, relative
velocities)
Coefficient of Static Friction on
Dry Surfaces
Materials
Coefficient of Static Friction or μ
Metal on metal
0.15 – 0.60
Metal on wood
0.20 – 0.60
Metal on stone
0.30 – 0.70
Metal on leather
0.30 – 0.60
Wood on wood
0.25 – 0.50
Wood on leather
0.25 – 0.50
Stone on stone
0.40 – 0.70
Earth on earth
0.20 – 1.00
Rubber on concrete
0.60 – 0.90
Friction
 Two types of friction

Dry or Coulomb friction
Static – surfaces are at rest with respect to each other
 Kinetic/dynamic – surfaces are moving with respect to
each other


Fluid Friction

Viscosity, friction developed between layers of a fluid
moving at different velocities
We’ll limit our discussion to dry friction
Friction Theory
 Consider a block on a horizontal surface


No Frictional resistance
N is referred to as the Normal force, perpendicular to the
contact surface
W
N
Friction Theory
 Next an external force is applied



If no friction, block would immediately move
If force is applied, F = P, for equilibrium
At some point, P > F, motion occurs
W
P
F
N
Friction Theory
Ratio of F to N is called the Coefficient of Static
Friction or μ
Where
F
s 
N
μs = coefficient of static friction (unitless)
F – Max or limiting frictional force resistance (lb, k, N)
N – Normal force, perpendicular to contact surface (lb, k, N)
Friction Theory
Stated another way:
Fs = μs N
(for impending motion)
Fk = μk N
(for sustaining motion, kinetic)
In general:
F≤
μN
Example Problem
Given:
W = 400 lb.
μ = 0.4
Find:
Force P required to
cause motion to
impend
W
P
Example Problem Solution
W
P
Given:
W = 400 lb.
μ = 0.4
Find:
P for impending motion
Solution:
1-FBD
2-Eqn of Equil
F=μN
F
y
0
N  400  0
N  400 lb 
400 lb
P
N
F
F  N
F  0.4  400
F  160 lb 
F
0
PF 0
P 160  0
P  160 lb 
x